# Electrical Circuit Question

1. Jan 24, 2013

### GreenPrint

1. The problem statement, all variables and given/known data

So I was wondering if someone could help with part C. I have gotten the answers for A and B, at least I hope they are correct.

2. Relevant equations

3. The attempt at a solution

So I'll just do the rectangle first and then try and solve the trapezoid. By the way do you think it's a triangle or a trapezoid? To me it loos like a trapazoid with a value at point l=0 of (0,.1) and not (0,0)???

Anyways for the rectangle...

So I'm given

rho, electron concentration - n, the thickness of the graphite layer - h (found in part a), and the velocity - v (I found it), and the voltage across the rectangle is 2 volts.

I'm confused as to how to find the voltage across the length. I thought that the voltage across the length would be a constant 2 volts but I guess I'm wrong.

Thanks for any help anyone can provide.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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Last edited: Jan 24, 2013
2. Jan 24, 2013

### Staff: Mentor

Voltage is 0 at the left side, and does not make jumps.
Current is constant, and resistance per length is constant. What can you tell about the voltage drop per length based on that?

3. Jan 24, 2013

### GreenPrint

I'm not exactly sure. If you rearrange the formula

R = rho(l/A)

And in this case indeed

R/l = rho/A

I can tell that from resistance per length is constant

from ohms law

V = I R

divide both sides by L

V/L = (I R)/L

V/L = I (rho/A)

so then Voltage as a function of length...

V(L) = (L*I*rho)/A

Now for I... I just use the fact that the resistance is given, 100k ohms and use 2 volts for the voltage

V = I*R

I = V/R

in the equation to find voltage as a function of length???

4. Jan 24, 2013

### Staff: Mentor

Well, you can use a shortcut: Once you know that voltage is a linear function, you can directly get the result from the boundary conditions: 0V at the left side and 2V at the right side.

5. Jan 24, 2013

### GreenPrint

http://img6.imageshack.us/img6/3449/capture6k.png [Broken]

Ok I guess I'm doing something wrong as my answer to C for the voltage makes no sense. Thanks for any help. I have shown all steps.

For part (a)
R = 100 kΩ
l = 10 cm
ρ = $10^{-5}$ Ωm

R = $\frac{ρl}{A}$
$\frac{1}{R} = \frac{A}{ρl}$
A = $\frac{ρl}{R} = \frac{10^{-5} Ωm(10 cm)kΩ(100 cm)}{(100 kΩ)(1000 Ω) m} = 10^{-7} cm^{2}$

A = wh, $h = \frac{A}{w} = \frac{10^{-7}cm^{2}}{cm} = 10^{-7} cm$

For part (b) trapezoid only

R = $\frac{ρl}{A}$, $\frac{1-.1}{10}l + .1 = .09 l + .1 = w$, A = wh, A(l) = (.09l + .1)$10^{-7} cm^{2}$
R(l) = $\frac{ρl}{A(l)}, R = \frac{10^{-5} Ωm (100 cm)}{10^{-7}cm^{2}m} ∫^{10}_{0}\frac{l dl}{.09l + .1}dl ≈ 471.889 Ω$

I = nqvA, I = $\frac{V}{R}, \frac{V}{R} = nqvA, v = \frac{V}{RnqA} = \frac{2 V}{471.889 Ω\frac{10^{19}}{cm^{3}}(1.6*10^{-19})(.09l+.1)(10^{-7}cm^{2})} = \frac{2648866.285}{.09l+.1} \frac{cm}{s}$

V = IR, I = $\frac{V}{R} = \frac{2V}{471.889 Ω} ≈ 4.238*10^{-3} A$

R = $\frac{lρ}{A}, \frac{R}{l} = \frac{ρ}{A}, V = IR, \frac{V}{l} = \frac{IR}{l} = \frac{Iρ}{A}$
$V(l) = \frac{lIρ}{A} = \frac{l4.238*10^{-3} A(10^{-5} Ωm)(100 cm)}{(.09l+.1)10^{-7}cm^{2}m} = \frac{l42.38}{.09l + .1}$

Last edited by a moderator: May 6, 2017
6. Jan 25, 2013

### Staff: Mentor

The last two lines are problematic:
$\frac{dV}{dl}=\frac{I\rho}{A}$
It is the derivative here.
$V(l)=\int_0^l \frac{dV}{dl} dl' = \int_0^l \frac{I\rho}{A(l')} dl'$
Where A depends on l at the trapezoid.

7. Jan 25, 2013

### GreenPrint

Wait so is the information you provide in your post correct?

V(l) = integral[0,l] [(I rho)/(A(l'))]dl'

8. Jan 25, 2013

### GreenPrint

Ok so for the cross sectional area of the trapezoid...
dA = h * w(l) * dl

So

A(l) = 10^(-7) integral[0,l] [.09*l+.1]dl

9. Jan 25, 2013

### Staff: Mentor

Do you have a reason to expect wrong formulas?

That (A in post 8) is not the same A you used above.
You don't need any integral to express A.

10. Jan 25, 2013

### GreenPrint

Oh thanks for the help by the way. I really appreciate it =)

Oh I guess I thought you were asking if it was right or something.

So the formula I had for A originally was correct?

11. Jan 25, 2013

### GreenPrint

So

$\frac{V}{l} = \frac{Iρ}{A(l)}$

$\frac{dV}{dl} = \frac{Iρ}{A}$

V(l) = $∫^{l}_{0} \frac{dV}{dl} dl = ∫^{l}_{0} dV = V$

I calculated earlier that

I = 4.238*$10^{-3} A$

and

A(l) = (.09*l + .1)$10^{-7}cm^{2}$

So

V(l) = $∫^{l}_{0} = \frac{4.238*10^{-3} A (10^{-5} Ωm (100 cm)}{(.09l + .1)10^{-7}cm^{2}m} dl$

I don't is right because when I plug in 10 I get about 423.8

12. Jan 25, 2013

### Staff: Mentor

Probably a numerical issue somewhere. Your resistance for the trapezoid is wrong - it has to be more than 100 kΩ (if you remove conductible material, the resistance increases), probably ~300-600 kΩ (471kΩ?)

13. Jan 25, 2013

### GreenPrint

Ok thanks, I think I found my mistake and got the correct answer. I actually didn't get 471k for the resistance of the trapezoid? Is that what I'm supposed to get? I did find that my answer for the resistance for the trapezoid was indeed wrong but that's not what I got when I found the correct answer to that. Well I don't know if it's correct but when I plugged it in to find the current and used that in my voltage equation and then plugged in 10 cm I got 2!

But my concern is that for the velocity of the square

I got something like

K/l

Where K is the constant that I found. If you plug in zero than the velocity is undefined. Is this a problem?

For the trapazoid I got something with a very well defined velocity at l = 0

I got something of the form

K/A(l)

When you plug in zero for L you don't get an undefined answer but a very well defined one. Is this an issue?

14. Jan 25, 2013

### Staff: Mentor

471kΩ was just a guess, based on your previous value of 471Ω. I did not calculate it.

The velocity in the square should be the same everywhere, you can calculate it with the known cross-section area, current and material constants, and they are all independent of l.

15. Jan 26, 2013

### GreenPrint

Thanks for all the help. I think I got the right answer (because of your help) =)